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leetcode-Remove Nth Node From End of List

时间:2014-11-27 09:14:31      阅读:197      评论:0      收藏:0      [点我收藏+]

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Given a linked list, remove the nth node from the end of list and return its head.

For example,

   Given linked list: 1->2->3->4->5, and n = 2.

   After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:
Given n will always be valid.
Try to do this in one pass.

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 Linked List Two Pointers
Have you met this question in a real interview?


/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        
        ListNode *p1,*p2;

        p1=head;
        p2=head;
        
        for(int i=1;i<=n;i++)
            p2=p2->next;
        
        if(p2==NULL)
        {
            p1=p1->next;
            return p1;
        }
            
        while(p2->next!=NULL)
        {
            p1=p1->next;
            p2=p2->next;
        }
        
        p1->next=p1->next->next;
        return head;
    }
};


leetcode-Remove Nth Node From End of List

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原文地址:http://blog.csdn.net/u013011841/article/details/41534789

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